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°.
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Q22. Let ABC be an acute-angled triangle with circumcenter O. Let P on BC be the foot of the altitude from A.
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[tex]\large\underline{\sf{Solution-}}[/tex]
➢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.
[tex]\rm :\longmapsto\:\angle AOB = 2\angle ACB[/tex]
and
[tex]\rm :\longmapsto\:\angle AOC = 2\angle ABC[/tex]
It is given that,
[tex]\rm :\longmapsto \: \angle BCA \geqslant \angle ABC + 30\degree [/tex]
On multiply by 2 on both sides, we get
[tex]\rm :\longmapsto \: 2\angle BCA \geqslant 2\angle ABC + 60\degree [/tex]
[tex]\rm :\longmapsto \: \angle AOB \geqslant \angle AOC + 60\degree [/tex]
[tex]\rm :\longmapsto \: \angle AOB \geqslant \angle ROB + 60\degree [/tex]
[ using (1) ]
[tex]\rm :\longmapsto \: \angle AOB - \angle ROB \geqslant 60\degree [/tex]
[tex]\rm :\longmapsto \: \angle AOR \geqslant 60\degree [/tex]
Now,
[tex]\rm :\longmapsto \: Consider \: \triangle AOR[/tex]
[tex]\rm :\longmapsto \: \angle AOR \geqslant 60\degree [/tex]
➢By using triangle inequality, we get
[tex]\rm :\implies\:AR \geqslant OR[/tex]
[tex]\rm :\implies\:AR \geqslant r[/tex]
[tex]\rm :\implies\:QP \geqslant r - - - (2)[/tex]
Now,
[tex]\rm :\longmapsto \: Consider \: \triangle QOC[/tex]
➢ We know, sum of two sides of a triangle is greater than third side.
So, using this,
[tex]\rm :\longmapsto\:OQ + OC > QC[/tex]
[tex]\rm :\longmapsto\:OQ + OC > QP + PC - - - (3)[/tex]
On comparing equation (2) and (3), we get
[tex]\rm :\longmapsto\:OP > PC[/tex]
We know, angle opposite to longest side is always greater
[tex]\rm :\implies\:\angle OCP > \angle COP - - - (4)[/tex]
Now,
[tex]\rm :\longmapsto\:In \: \triangle \: BOC[/tex]
[tex]\rm :\longmapsto\:OB \: = \: OC = r[/tex]
[tex]\rm :\longmapsto\:\angle OBC \: = \: \angle OCB [/tex]
Thus,
[tex]\rm :\longmapsto\:\angle BOC = 180\degree - 2\angle OCB[/tex]
can be rewritten as
[tex]\rm :\longmapsto\:\angle BOC = 180\degree - 2\angle OCP[/tex]
can be rewritten as
[tex]\rm :\longmapsto\:2\angle BAC = 180\degree -2 \angle OCP[/tex]
[tex]\rm :\longmapsto\:\angle BAC = 90\degree - \angle OCP[/tex]
[tex]\rm :\longmapsto\:\angle BAC \: + \: \angle OCP \: = \: 90\degree - - - (5)[/tex]
From equation (4) and (5) we concluded that
[tex]\rm :\longmapsto\:\angle BAC \: + \: \angle COP \: < \: 90\degree[/tex]
Answer:
referred to the attachment.