Mathematics
In the adjoining figure, ABCD is a parallelogram. BM ⟂ AC and DN ⟂ AC. Prove that :
(i) ΔBMC ≅ ΔDNA.
(ii) BM = DN.
Answer
(i) In triangle DNA and BMC,
∠DNA = ∠BMC = 90° (Given)
∠DAN = ∠MCB [Alternate interior angles BC and AD are parallel, AC acts as a transversal]
BC = AD [opposite sides of parallelogram]
∴ ΔBMC ≅ ΔDNA.[By A.A.S. rule]
Hence, proved that ΔBMC ≅ ΔDNA.
(ii) We know that,
ΔBMC ≅ ΔDNA
∴ BM = DN [Corresponding sides of Congruent Triangles]
Hence, proved that BM = DN.
Related Questions
If one angle of a parallelogram is 90°, show that each of its angles measures 90°.
In the adjoining figure, ABCD and PQBA are two parallelograms. Prove that :
(i) DPQC is a parallelogram.
(ii) DP = CQ.
(iii) ΔDAP ≅ ΔCBQ.
In the adjoining figure, ABCD is a parallelogram and X is the mid-point of BC. The line AX produced meets DC produced at Q. The parallelogram AQPB is completed. Prove that :
(i) ΔABX ≅ ΔQCX.
(ii) DC = CQ = QP.
In the adjoining figure, ABCD is a parallelogram. Line segments AX and CY bisect ∠A and ∠C respectively. Prove that :
(i) ΔADX ≅ ΔCBY
(ii) AX = CY
(iii) AX ∥ CY
(iv) AYCX is a parallelogram
