Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 8.1, 9 In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ Show that: Δ APD ≅ Δ CQB Given: ABCD is a parallelogram where DP = BQ To prove: Δ APD ≅ Δ CQB Proof: Now, AD ∥ BC and transversal BD. ∠ADP = ∠CBQ In ΔAPD and ΔCQB, AD = CB ∠ADP = ∠CBQ DP = BQ ∴ ΔAPD ≅ ΔCQB Ex 8.1, 9 In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ Show that: (ii) AP = CQ In previous part, we proved that ΔAPD ≅ ΔCQB, ∴ AP = CQ Ex 8.1, 9 In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ Show that: (iii) Δ AQB ≅ Δ CPD Since, AB ∥ DC and transversal BD. ∠ABQ = ∠CDP In ΔAQB and ΔCPD, AB = CD ∠ABQ = ∠CDP BQ = DP ∴ ΔAQB ≅ ΔCPD Ex 8.1, 9 In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ. Show that: (iv) AQ = CP In previous part, we proved that ΔAQB ≅ ΔCPD, ∴ AQ = CP Ex 8.1, 9 In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ. Show that: (v) APCQ is a parallelogram In part (ii) & (iv) we proved that AP = CQ and AQ = CP Since both pairs of opposite sides in APCQ are equal, APCQ is a parallelogram.

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.