Question

Q22. Let ABC be an acute-angled triangle with circumcenter O. Let P on BC be the foot of the altitude from A. Suppose that ∠BCA ≥ ∠ABC + 30°. Prove that, ∠CAB + ∠COP < 90°.​

Answer 1

$\large\underline{\sf{Solution-}}$

➢Let assume that 'l' draw perpendicular bisector of BC. Let further assume that reflection of points A and P under 'l' be R and Q.

➢So, APQR is a rectangle, and BQ = PC and triangle RQB is congruent to triangle ACB (By RHS Congruency Rule)

So, RB = AC.

We know,

➢Equal chords subtends equal angles at the centre.

∴ ∠ROB = ∠AOC-----(1).

Now, also we know that,

➢Angle subtended at the centre is twice the angle subtended at the circumference by the same arc.

$\rm :\longmapsto\:\angle AOB = 2\angle ACB$

and

$\rm :\longmapsto\:\angle AOC = 2\angle ABC$

It is given that,

$\rm :\longmapsto \: \angle BCA \geqslant \angle ABC + 30\degree$

On multiply by 2 on both sides, we get

$\rm :\longmapsto \: 2\angle BCA \geqslant 2\angle ABC + 60\degree$

$\rm :\longmapsto \: \angle AOB \geqslant \angle AOC + 60\degree$

$\rm :\longmapsto \: \angle AOB \geqslant \angle ROB + 60\degree$

[ using (1) ]

$\rm :\longmapsto \: \angle AOB - \angle ROB \geqslant 60\degree$

$\rm :\longmapsto \: \angle AOR \geqslant 60\degree$

Now,

$\rm :\longmapsto \: Consider \: \triangle AOR$

$\rm :\longmapsto \: \angle AOR \geqslant 60\degree$

➢By using triangle inequality, we get

$\rm :\implies\:AR \geqslant OR$

$\rm :\implies\:AR \geqslant r$

$\rm :\implies\:QP \geqslant r - - - (2)$

Now,

$\rm :\longmapsto \: Consider \: \triangle QOC$

➢ We know, sum of two sides of a triangle is greater than third side.

So, using this,

$\rm :\longmapsto\:OQ + OC > QC$

$\rm :\longmapsto\:OQ + OC > QP + PC - - - (3)$

On comparing equation (2) and (3), we get

$\rm :\longmapsto\:OP > PC$

We know, angle opposite to longest side is always greater

$\rm :\implies\:\angle OCP > \angle COP - - - (4)$

Now,

$\rm :\longmapsto\:In \: \triangle \: BOC$

$\rm :\longmapsto\:OB \: = \: OC = r$

$\rm :\longmapsto\:\angle OBC \: = \: \angle OCB$

Thus,

$\rm :\longmapsto\:\angle BOC = 180\degree - 2\angle OCB$

can be rewritten as

$\rm :\longmapsto\:\angle BOC = 180\degree - 2\angle OCP$

can be rewritten as

$\rm :\longmapsto\:2\angle BAC = 180\degree -2 \angle OCP$

$\rm :\longmapsto\:\angle BAC = 90\degree - \angle OCP$

$\rm :\longmapsto\:\angle BAC \: + \: \angle OCP \: = \: 90\degree - - - (5)$

From equation (4) and (5) we concluded that

$\rm :\longmapsto\:\angle BAC \: + \: \angle COP \: < \: 90\degree$

Answer 2

Answer:

referred to the attachment.