# In a triangle ABC, if angle A =60 and altitudes from B and C meet AC and AB at P and Q respectively and intersect each other at I. Prove that APIQ and PQBC are cyclic quadrilaterals.

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In quadrilateral AQIP,

∠A=60 degree

∠AQI=∠API=90 degree

So ∠PIQ=360−90−90−60=120 degree

So we can see that in quadrilateral AQIP ,

∠A+∠PIQ=180 and ∠AQI+∠API=180

SO APIQ is a cyclic quadrilateral.

Now in triangle ABP, angle APC is a external angle

So 90=60+∠1

∠1=30 degree

and in triangle AQC, angle BQC is a external angle so

90=60+∠2

∠2=30 degree

SO ∠1=∠2

So we can see that it is only possible when PQBC are cyclic qudrilateral.

∠A=60 degree

∠AQI=∠API=90 degree

So ∠PIQ=360−90−90−60=120 degree

So we can see that in quadrilateral AQIP ,

∠A+∠PIQ=180 and ∠AQI+∠API=180

SO APIQ is a cyclic quadrilateral.

Now in triangle ABP, angle APC is a external angle

So 90=60+∠1

∠1=30 degree

and in triangle AQC, angle BQC is a external angle so

90=60+∠2

∠2=30 degree

SO ∠1=∠2

So we can see that it is only possible when PQBC are cyclic qudrilateral.