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$$PQR$$ is an isosceles triangle where $$PQ = PR$$. $$X$$ is a point on the circumcircle of $$\triangle PQR$$, such that it being in the opposite region of $$P$$ with respect to $$QR$$. The normal drawn from the point $$P$$ on $$XR$$ intersects $$XR$$ at point $$Y$$. If $$XY = 10$$, then find the value of $$QX + RX$$.

Note by Ashraful Mahin
11 months ago

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- 5 months, 2 weeks ago

its easy. extend XR to a point M such that RM=QX. Now try proving that Y is the mid point of XM.

- 8 months, 2 weeks ago

What have you tried?

Have you drawn a diagram? If so, what does it look like?

Staff - 11 months ago