# Category: mc9cs_anglestriangles

## ABCD is a Rhombus. DPR and CBR are straight lines. Prove that: DP.CR=DC.PR

ABCD is a rhombus. DPR and CBR are straight lines. Prove that: DP.CR=DC.PR Solution: DP.CR=DC.PR Given ABCD is rhombus . DPR and CBR are straight lines.∴ AD||CR we need to Prove : DP.CR=DC.PRIn ∆ ADP and ∆ PCRWe have : ∠ APD = ∠ CPR ∠ ADP = ∠ PRC ∠ DAP = ∠ PCR…

## In triangle ABC, AB > AC. | E is the mid-point of BC | and AD is perpendicular to BC

In triangle ABC, AB > AC. E is the mid-point of BC and AD is perpendicular to BC. Prove that AB2 + AC2 = 2.AE2 + 2.BE2 Solution:In ∆ ABD AB2 = AD 2 + BD 2 Or AB2 = AD 2 + (BE + ED ) 2 Or AB2 = AD 2 + BE…

## OP, OQ and OR are drawn perpendiculars to the sides BC, CA and AB of ∆ ABC |Prove that AR2 +BP2 +CQ2 = AQ 2 + CP 2+BR2

In the following figure, OP, OQ and OR are drawn perpendiculars to the sides BC, CA and AB respectively of triangle ABC. Prove that AR2 +BP2 +CQ2 = AQ 2 + CP 2+BR2 solution : Construction : joinA-0, BO and CO . For ∆ OAR, we can write:AR2 = OA2 – OR2 —— Eqn. (I)For…

## AB = BC and AD is perpendicular to CD. Prove that AC2 = 2.BC. DC

Question:In the figure AB = BC and AD is perpendicular to CD. Prove that AC2 = 2.BC. DC solution :In ∆ ADC AC2 = AD2 + DC2 AC2 = AB2 – BD2 + DC2(AS ABD is a right angled triangle)or AC2 = AB2 + DC2 – BD2 AC2 = AB2 + DC2 – ( DC-BC…

## ABCD is a parallelogram | Prove that ∆ BCF and ∆ DFE are equal in area

ABCD is a parallelogram , a line trhough A cuts DC at oint F and BC produced at E Prove that ∆ BCF is equal in area to ∆ DFE. Construction: AC is joined. AB || CD and AD || BC .Now ∆ ACE and ∆ DCE are on the same base CE and between…

## Geometry Sum | ABCD is a square. Prove in each case ∆APD ≅ ∆BPC | Find Angle DPC Angle PCD and Angle PCD |

ABCD is a square and APB is an equilateral triangle. Prove in each case: 1) 𢁪PD ≅ 𢁫PC 2) Find the angles of 𢁭PC Solution : First PartGiven �P = Equilateral triangleR.T.P.1) 𢁪PD ≅ 𢁫PC 2) Find 𢈍PC, ∠PDC and ∠PCD, Proof: In 𢁪PD and 𢁫PC AP = BP | Same side of…