Question 1101828: ABC is a right angle isosceles triangle, angle BCA = 90, with BC as the base and AB as hypotenuse side AC = BC. Point M is at midpoint on hypotenuse such that BM=MA=20cm. P and Q are points on sides BC and AC respectively. A equilateral triangle is formed by joining MPQ. Find the length, in CM, of segment PC.
Found 2 solutions by Boreal, greenestamps:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Draw this.
Triangles AMQ and BPM are similar.
Both have an angle of 45 degrees (this is an isosceles right triangle, so it is 45-45-90), they have a side of the equilateral triangle, and the angle PM and QM make with the hypotenuse must be 60 degrees, since the triangle is equilateral and the two other angles must be equal to each other. That makes the third angle in those triangles 75 degrees.
Now we have 3 angles, and the side opposite to the 75 degree angle is 20
Law of sines where sin 75/20=sin 60/BP, and BP is 17.93 cm.
The side BC is 40/sqrt(2)=20 sqrt(2)
20 sqrt(2)-17.93=10.35 cm.
I can check this by looking at the equilateral triangle, which with the Law of Sines has sides of 14.64 cm.
Then, right triangle PCQ is a 45-45-90 triangle, with the hypotenuse 14.64 cm. Each of the two legs is equal and is 10.35 cm, which is 14.64/sqrt(2)
Answer by greenestamps(13195)  (Show Source):
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