Question

ABC is a triangle in which AB=AC and P is a point on AC Through C a line is drawn to intersect BP produced at Q such that angle ABQ=angle ACQ prove that angle ACQ=90° +1/2 angle BAC ​

Answer 1

Answer:

proved

Step-by-step explanation:

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

Answer 2

Answer:

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