Question

ABCD is a rhombus whose side AB is produced to points P and Q such that AP=AB=BQ. PD and QC are produced to meet at a point R. Show that

Answer 1

Given PA = AB = BQ,
We know AB = CD = BC = AD (sides of rhombus are equals)
And ∠DOC = ∠AOD = ∠AOB = ∠BOC = 90° ( Diagonal of rhombus are perpendicular to each other)
In ∆PAD, We know
PA = AD
So,
∠APD = ∠ADP = x (Angle opposites same side are equal As PA = AD)
Then
∠PAD = 180° - x
Similarly In ​ ∆BCQ
BC = BQ
So,
∠BCQ = ∠BQC = y (Angle opposites same side are equal As PA = AD)
Then
∠QBC = 180° - y
So, In ∆ RDC
∠RDC = ∠APD = x (As we know Two lines are parallel if and only if the two angles of any pair of corresponding angles of any transversal are congruent)
Therefore
∠BQC = ∠RCD = y
And,
∠PAD = ∠ADC = 180° - x (As we know Two lines are parallel if and only if the two angles of any pair of alternating angles of any transversal are congruent)
Therefore
∠QBC​ = ∠BCD = 180° - y
Now,
∠CDO = 12 ∠ADC = 90° - x ( As we know diagonal of rhombus bisect its angles)
Therefore
∠DCO = 12 ∠BCD = 90° - y
So In quadrilateral RDOC
∠DOC = 90°
∠RDO = ∠CDO + ∠RDC
= 90° - x + x
= 90°
∠RCO = ​∠DCO + ​∠RCD
= 90° - y + y
= 90°
SO the remainig ​∠DRC = 90° (Because sum of internal angles of any quadrilateral is 180°)
Therefor
∠PRQ = 90° (Because ∠PRQ = ​∠DRC )
Hence ​∠PRQ is a right angle (Hence proved)
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Answer 2

answer which is up is absolutely right