# Prove it

Suppose that P and Q are points on the sides AB and AC respectively of △ABC. The perpendiculars to the sides AB and AC at P and Q respectively meet at D, an interior point of △ABC. If M is the midpoint of BC, prove that PM = QM if and only if ∠BDP=∠CDQ .

I tried to come up with a solution using Euclidean geometry only, but to no avail.

Note by Abu Nazam Sk Md Tamimuddin
1 month, 1 week ago

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$PM=QM$ is a direct result of the Nine-Point Circle Theorem

- 1 month ago

But P and Q are given to be random points on AB & AC respectively. And yes, the result holds when D is the orthocentre.

- 1 month ago

But points P and Q are given to be random which indicates that points P, D and C should not be collinear necessarily. Same thing applies for points Q, D and B.

Yes, but when the angles are equal, doesn't it imply that they are collinear (vertically opposite angles) ?

- 1 month ago

No, not necessarily. It's still possible for them to be equal without being intersected in lines.

I found a solution here. https://math.stackexchange.com/questions/3961588/geometric-problem-concerning-relation-of-equal-segments-and-angles

- 3 weeks, 3 days ago

That was a nice problem!

- 3 weeks, 3 days ago

Indeed!

- 3 weeks, 1 day ago