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
7 months, 1 week ago

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

Sathvik Acharya - 7 months ago

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

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Sathvik Acharya - 7 months ago

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@Sathvik Acharya 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.

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@Abu Nazam Sk Md Tamimuddin Yes, but when the angles are equal, doesn't it imply that they are collinear (vertically opposite angles) ?

Sathvik Acharya - 7 months ago

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

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

Abu Nazam Sk Md Tamimuddin - 6 months, 3 weeks ago

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That was a nice problem!

Sathvik Acharya - 6 months, 3 weeks ago

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Indeed!

Abu Nazam Sk Md Tamimuddin - 6 months, 3 weeks ago

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test

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I tested the result through a software. It holds.

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