Question

ABC is a triangle in which AB=AC and P is any point on AC. Through C a line is drawn to intersect BP produce at Q such that angle ABQ=ACQ PROVE THAT ANGLE AQC =90+1÷2anleBAC​

Answer 1

Answer:

proved

$See attached figure</p><p>Since ABD and ACD are angles in the same segment of chord AD, points A, B, C and D have to lie on the circle, i.e., they have to be concyclic.</p><p>AB = AC</p><p>∠ABC = ∠ACB</p><p>In triangle ABC,</p><p>∠ABC + ∠BAC + ∠ACB = 180°</p><p>∴∠BAC + 2∠ACB = 180°</p><p>2∠ACB = 180 - ∠BAC</p><p>∠ACB = \frac{180 - BAC}{2}2180−BAC</p><p>∠ACB = 90 - \frac{1}{2}21 BAC - - - -(i)</p><p>∠ACB = ∠ADB (ii) ∵Angles in the same segment</p><p>And ∠BAC = ∠BDC (iii) ∵Angles in the same segment</p><p>Adding (ii) and (iii)</p><p>∠ACB + ∠BAC = ∠ADB + ∠ BDC</p><p>∠ACB + ∠BAC = ∠ADC</p><p>∠ACB = ∠ADC - ∠BAC (iv)</p><p>Plugging value of ∠ACB from (i) in (iv),</p><p>∠ADC - ∠BAC = 90 - \frac{1}{2}21 BAC</p><p>∠ADC = 90 - \frac{1}{2}21 BAC + 1∠BAC</p><p>∠ADC = 90 + \frac{1}{2}21 BAC</p><p>==================</p><p>Hence proved.</p><p>$

Step-by-step explanation:

See attached figure

Since ABD and ACD are angles in the same segment of chord AD, points A, B, C and D have to lie on the circle, i.e., they have to be concyclic.

AB = AC

∠ABC = ∠ACB

In triangle ABC,

∠ABC + ∠BAC + ∠ACB = 180°

∴∠BAC + 2∠ACB = 180°

2∠ACB = 180 - ∠BAC

∠ACB = \frac{180 - BAC}{2}2180−BAC

∠ACB = 90 - \frac{1}{2}21 BAC - - - -(i)

∠ACB = ∠ADB (ii) ∵Angles in the same segment

And ∠BAC = ∠BDC (iii) ∵Angles in the same segment

Adding (ii) and (iii)

∠ACB + ∠BAC = ∠ADB + ∠ BDC

∠ACB + ∠BAC = ∠ADC

∠ACB = ∠ADC - ∠BAC (iv)

Plugging value of ∠ACB from (i) in (iv),

∠ADC - ∠BAC = 90 - \frac{1}{2}21 BAC

∠ADC = 90 - \frac{1}{2}21 BAC + 1∠BAC

∠ADC = 90 + \frac{1}{2}21 BAC

==================

Hence proved.