CPCTC - Math 1

In right triangle $ABC$ with right angle at $C$ , the altitude $\overline{CD}$ is drawn to hypotenuse $\overline{AB}$ . If $\triangle ACD \cong \triangle CBD$ , what can be concluded about triangle $ABC$ ?
Answer: Triangle $ABC$ is isosceles with $AC = BC$ When you encounter a problem involving congruent triangles formed by an altitude to the hypotenuse of a right triangle, focus on what the congruence tells you about the original triangle's properties. Given that $\triangle ACD \cong \triangle CBD$ , these triangles share the altitude $\overline{CD}$ and both contain right angles ( $\angle ADC = \angle BDC = 90°$ ). For the triangles to be congruent, their corresponding sides must be equal. Since $CD$ is common to both triangles, we need $AC = BC$ for congruence to hold. This makes triangle $ABC$ isosceles with the two legs equal. You can verify this by noting that if $AC = BC$ , then $\angle CAD = \angle CBD$ (the base angles of the isosceles triangle), and the triangles become congruent by AAS (Angle-Angle-Side). Answer A is incorrect because while the triangle has equal base angles, they're each $45°$ , not $30°$ and $60°$ . In an isosceles right triangle, the two acute angles are always $45°$ each. Answer B is wrong because it states $AB = AC$ . Since $AB$ is the hypotenuse and $AC$ is a leg, this would violate the fundamental property that the hypotenuse is the longest side in a right triangle. Answer D is impossible since we already have a $90°$ angle at $C$ , so the triangle cannot have all $60°$ angles. Study tip: When an altitude creates congruent triangles in a right triangle, the original triangle is always isosceles with equal legs, making it a $45°$ - $45°$ - $90°$ triangle.
Quadrilateral $PQRS$ has diagonals that bisect each other at point $T$ . If $\triangle PQT \cong \triangle RST$ , which statement must be true by CPCTC?
Answer: $\overline{PQ} \cong \overline{RS}$ and $\overline{PS} \cong \overline{QR}$ Since $\triangle PQT \cong \triangle RST$ , by CPCTC we know that $\overline{PQ} \cong \overline{RS}$ and $\overline{PS} \cong \overline{QR}$ (noting that $\overline{PT} \cong \overline{RT}$ from the given congruence). Choice A describes parallelism, which requires additional reasoning beyond CPCTC. Choice C states what we already know from the diagonals bisecting each other. Choice D describes vertical angles, which are always congruent regardless of triangle congruence.
In parallelogram $WXYZ$ , diagonal $\overline{WY}$ is drawn. Given that $\triangle WXY \cong \triangle YZW$ , and $m\angle XWY = 35°$ , what is $m\angle WYZ$ ?
Answer: $35°$ When you encounter congruent triangles in a parallelogram with a diagonal, you're dealing with corresponding angles and the properties of parallel sides. Given that $\triangle WXY \cong \triangle YZW$ , the key insight is identifying which angles correspond to each other. In these congruent triangles, angle $XWY$ in the first triangle corresponds to angle $ZYW$ in the second triangle. Since corresponding angles in congruent triangles are equal, and we know $m\angle XWY = 35°$ , we can conclude that $m\angle ZYW = 35°$ . But wait - angle $ZYW$ is the same as angle $WYZ$ ! They're just written in different notation. So $m\angle WYZ = 35°$ , making D correct. Let's examine why the other answers are wrong. Choice A ( $145°$ ) likely comes from incorrectly thinking this angle is supplementary to the given $35°$ angle, but these angles aren't supplementary. Choice B ( $55°$ ) might result from subtracting $35°$ from $90°$ , perhaps assuming a right angle exists where none is given. Choice C ( $70°$ ) could come from doubling the given angle, which has no geometric basis in this problem. Study tip: When working with congruent triangles, always carefully match up corresponding vertices in the congruence statement. The order matters - $\triangle WXY \cong \triangle YZW$ tells you that W↔Y, X↔Z, and Y↔W, which directly gives you the angle correspondences you need.
Two triangles $\triangle JKL$ and $\triangle MNO$ are congruent. If $J$ corresponds to $M$ , $K$ corresponds to $N$ , and $L$ corresponds to $O$ , and if $m\angle J = 4y + 10$ , $m\angle M = 6y - 20$ , $m\angle K = 3z + 5$ , what is the value of $z$ when $m\angle N = 65°$ ?
Answer: $z = 20$ Since the triangles are congruent with the given correspondence, by CPCTC we have $\angle J \cong \angle M$ and $\angle K \cong \angle N$ . From $\angle J \cong \angle M$ : $4y + 10 = 6y - 20$ , so $30 = 2y$ and $y = 15$ . From $\angle K \cong \angle N$ : $3z + 5 = 65$ , so $3z = 60$ and $z = 20$ . Choice A results from solving $4y + 10 = 65$ incorrectly. Choice C results from $z = (65 + 10)/3$ . Choice D results from $z = (65 + 25)/3$ .
In triangles $ABC$ and $DEF$ , $\overline{AB} \cong \overline{DE}$ , $\overline{BC} \cong \overline{EF}$ , and $\angle B \cong \angle E$ . If $m\angle A = 3x + 15$ and $m\angle D = 5x - 5$ , what is the value of $x$ ?
Answer: $x = 10$ By SAS, triangles ABC and DEF are congruent. By CPCTC, corresponding angles are congruent, so $\angle A \cong \angle D$ . Therefore $3x + 15 = 5x - 5$ , which gives $20 = 2x$ , so $x = 10$ . Choice B results from solving $3x + 15 = 5x + 5$ (sign error). Choice C results from setting up $3x - 15 = 5x - 5$ (sign error in first expression). Choice D results from incorrectly solving $3x + 15 = 5x - 5$ as $15 + 5 = 5x - 3x$ , getting $20 = 2x$ but then making an arithmetic error.
Given that $\triangle PQR \cong \triangle STU$ by AAS, where $PQ = 2x - 3$ , $ST = x + 7$ , and $QR = 3y + 1$ . If $TU = 19$ , what is the value of $xy$ ?
Answer: 60 From the congruence statement △PQR ≅ △STU, corresponding sides are equal by CPCTC. PQ corresponds to ST, so 2x - 3 = x + 7, which gives x = 10. QR corresponds to TU, so 3y + 1 = 19, which gives 3y = 18, so y = 6. Therefore xy = 10 × 6 = 60.

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