SOLUTION: In right triangle ABC, AC = BC and ∠C = 90. Let P and Q be points on hypotenuse
AB, such that ∠PCQ = 45°. Show that AP^2+BQ^2=PQ^2. I think the first step is
to rotate it 90
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Triangles
-> SOLUTION: In right triangle ABC, AC = BC and ∠C = 90. Let P and Q be points on hypotenuse
AB, such that ∠PCQ = 45°. Show that AP^2+BQ^2=PQ^2. I think the first step is
to rotate it 90
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Question 1153572: In right triangle ABC, AC = BC and ∠C = 90. Let P and Q be points on hypotenuse
AB, such that ∠PCQ = 45°. Show that AP^2+BQ^2=PQ^2. I think the first step is
to rotate it 90 degrees counter clockwise around C, let P go to P’ and find ∠P’CQ. Found 3 solutions by josgarithmetic, Edwin McCravy, AnlytcPhil:Answer by josgarithmetic(39617) (Show Source):
You can put this solution on YOUR website! Notice the spaces added between your use of 'angle' symbol and character which follows them. Without each space, the problem description was not being fully displayed.
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In right triangle ABC, AC = BC and < C = 90. Let P and Q be points on hypotenuse AB , such that < PCQ = 45. Show that AP^2+BQ^2=PQ^2. I think the first step is to rotate it 90 degrees counter clockwise around C, let P go to P’ and find < P’CQ.
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[My previous solution had some lettering that did not match the drawn figure. I think I have corrected all the errors below.]
Draw lines CR and RB so that ΔCRB ≅ ΔCPA
Now draw in QR:
You can finish now. It's mostly corresponding parts of
congruent triangles and the Pythagorean theorem. Here are
some of the steps:
ΔCRB ≅ ΔCPA
BR = AP
∠ACP = ∠BCR
∠ACP + ∠PCQ + ∠QCB = 90°
∠ACP + 45° + ∠QCB = 90°
∠ACP + ∠QCB = 45°
∠BCR + ∠QCB = 45°
Show that ΔQCP ≅ ΔQCR
Then PQ = QR
≅QBR is a right triangle
BR² + BQ² = QR²
AP² + BQ² = PQ²
If you have trouble finishing, tell me in the space below,
and I'll get back to you be email. (No charge, ever!!)
Edwin
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