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.

#Geometry

Easy Math Editor

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Comments

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

Sathvik Acharya - 5 months, 3 weeks 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.

Abu Nazam Sk Md Tamimuddin - 5 months, 3 weeks ago

Sathvik Acharya - 5 months, 3 weeks ago

@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.

Abu Nazam Sk Md Tamimuddin - 5 months, 3 weeks ago

@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 - 5 months, 3 weeks ago

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

Abu Nazam Sk Md Tamimuddin - 5 months, 3 weeks ago

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 - 5 months, 2 weeks ago

That was a nice problem!

Sathvik Acharya - 5 months, 2 weeks ago

Indeed!

Abu Nazam Sk Md Tamimuddin - 5 months, 2 weeks ago

test

Banashri Banaja Laxmi Mohanty - 5 months, 3 weeks ago

I tested the result through a software. It holds.

Abu Nazam Sk Md Tamimuddin - 5 months, 3 weeks ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$