Question

In right triangle ABC on hypotenuse AB lies points K and M such that AK=AC, BM=BC. Find m∠MCK.

Answer 1

Answer:

The measure of ∠MCK is 45°

Step-by-step explanation:

The parameters given are;

ΔABC with hypotenuse side AB

Let AB = x

BC = y

AC = z

∴ x² = y² +  z²

∠BCM = ∠CMB  (Base angles of isosceles triangle ΔBMC)

Similarly, ∠ACK = ∠AKC (Base angles of isosceles triangle ΔKAC)

∴ ∠BCM + ∠CMB + ∠CBM = 180° (Sum of interior angles in a triangle)

Hence, 2 × ∠BCM + ∠CBM = 180°........(1)

Similarly, 2 × ∠ACK + ∠CAK = 180°......(2)

Hence, 2 × ∠BCM + ∠CBM = 2 × ∠ACK + ∠CAK

However, ∠CBM + ∠CAK + 90° = 180° (Sum of angles in right triangle ΔABC)

∴∠CBM + ∠CAK = 180° - 90° = 90°

Adding equations (1) and (2), we have;

2 × ∠BCM + ∠CBM + 2 × ∠ACK + ∠CAK = 180° + 180° = 360°

Which gives;

2 × ∠BCM + 2 × ∠ACK + ∠CBM + ∠CAK = 360°

Where;

∠CBM + ∠CAK = 90°, we have;

2 × ∠BCM + 2 × ∠ACK + 90° = 360°

∴ 2 × ∠BCM + 2 × ∠ACK = 360° - 90° = 270°

∠BCM + ∠ACK = 270°/2 = 135°

∠BCM = ∠CMB

∴ ∠CMB + ∠ACK = 135°

Also, ∠CMB = ∠CMK and ∠ACK = ∠MCK

Hence, ∠CMK + ∠MCK  = 135°

However, ∠CMK + ∠MCK + ∠MCK = 180° (Sum of interior angles in a triangle)

Which gives;

∠CMK + ∠MCK + ∠MCK = 135° + ∠MCK = 180°

∴ m∠MCK = 180° - 135° = 45°.

The measure of ∠MCK, m∠MCK = 45°.