What name shall I give to this problem?

Geometry Level 3

Let $\Delta ABC$ be a right angled triangle with $\angle C = 90^{o}$ . Suppose that $AP$ and $BQ$ be the angular bisectors of $\angle CAB$ and $\angle CBA$ respectively. Let $M$ and $N$ be perpendiculars respectively from $P$ and $Q$ on to $AB$ . Find the value of $\angle MCN$ (in degree).

This problem is from the set What name should I give?

The answer is 45.

1 solution

Surya Prakash
Aug 22, 2015

Observe that $\angle PMA + \angle PCA = 180^{o}$ . So, $P$ , $C$ , $A$ and $M$ are cyclic. This implies that $\angle MCP = \angle MAP$ .

Similarly, $\angle NCQ = \angle NBQ$ .

So, $\angle MCN = 90^{o} - \angle BCM - \angle ACN$ $= 90^{o} - \angle ABQ - \angle BAP$ $= 90^{o} -\dfrac{1}{2} \angle BAC - \dfrac{1}{2} \angle CBA$ $= 90^{o} - \dfrac{1}{2}(\angle BAC + \angle CBA)$

But, $\angle BAC + \angle CBA = 90^{o}$ . So,

$\angle MCN = \boxed{45^{o}}$

Moderator note:

Good observation with the cyclic quadrilaterals that help with the angle chasing.

The angle bisectors have been named wrong

Amartya Anshuman - 5 years, 9 months ago

Thanks for noting out.

Surya Prakash - 5 years, 9 months ago

How angle NCQ=NBQ ??

Chirayu Bhardwaj - 5 years, 3 months ago

What name shall I give to this problem?

This problem is from the set What name should I give?

The answer is 45.

1 solution

Moderator note:

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