Q. In fig ABCD is rhombus whose AB is produced to points P & Q such that AP=AB=BQ, PD & QC are producedto meet at a point R. Show that angle PRQ=90 °.



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Given; AP = AB = BQ; - 1
           ABCD is a rhombus;
So AB = BC = CD = AD;
From 1,

AB = BC = CD = AD = AP = AQ; - 2
Drawing diagonals AC and BD;

In triangle DPA,

DA = PA (proved earlier);
So angle DPA = angle PDA (angles opp. equal sides are equal);

From joining diagonals AC and BD,
In triangle ADB,
AD = AB (given earlier);
So angle ADB = angle ABD (angles opp. equal sides are equal);

In triangle BDC,
DC = BC (given earlier);
So angle BDC = angle CBD (angles opp. equal sides are equal);

And angle QDC = angle DPA (corresponding angles, opp. sides of a rhombus are parallel, extending AB and DC on both sides, taking DP as transversal);

From 2,
Angle PDA = angle ADB = angle BDC = angle CDQ; - 3

And angle PDA + angle ADB + angle BDC + angle CDQ = 180°;

So 4*angle PDA = 180°;

Angle PDA and all angles in 3 are 45°;

Like the procedure done above for finding angle CDQ, find angle DCQ;

So in triangle QDC,
Angle DQC + angle QDC + angle QCD = 180° (Angle sum property);

But angle QDC, angle QCD = 45°;

So 90° + angle DQC = 180°;

So angle DQC = angle PQR = 180° - 90° = 90° hence proved;

Hope this helps;

Cheers!

P.S - The triangle is not PQQ it should be PQR.....