Question

Let ABC be a right-angled triangle with ∠B = 90◦. Let BD be the altitude from B on to AC. Let P, Q and I be the incentres of triangles ABD, CBD and ABC respectively. Show that the circumcentre of of the triangle P IQ lies on the hypotenuse AC.

Answer 1

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Asked on December 26, 2019 byRoshna Mondal

Let ABC be a right-angled triangle with ∠B=90. Let BD be the altitude from B on to AC. Let P,Q and I be the incentres of triangles ABD,CBD, and ABC, respectively. Show that the circumcentre of the triangle PIQ lies on the hypotenuse AC.

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ANSWER

Solution 1:

We begin with the following lemma:

Lemma: Let XYZ be a triangle with ∠XYZ=90+α. Construct an isosceles triangle XEZ, externally on the side XZ, with base angle.

Then E is the circumcentre of ΔXYZ.

Proof of the Lemma: Draw ED ⊥XZ.

Then DE is the perpendicular bisector of XZ. We also observe that ∠XED=ZED=90−α .

 Observe that E is on the perpendicular bisector of XZ.

Construct the circumcircle of XY Z. Draw perpendicular bisector of XY and let it meet DE in F.

Then F is thecircumcentre of ΔXYZ. Join XF.

Then ∠XFD=90−α.

But we know that ∠XED=90−α. Hence E = F.

Let r1,r2 and r be the inradii of the triangles ABD, CBD and ABCrespectively. Join PD and DQ. Observe that ∠PDQ=90o. Hence

PQ2=PD2+DQ2=2r12+2r22:

Let s1 = (AB + BD + DA)=2. Observe that BD = ca=b and AD = AB2−BD2=c2−(bca