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 &#x2206 ADP and &#x2206 PCRWe have : &#x2220 APD = &#x2220 CPR &#x2220 ADP = &#x2220 PRC &#x2220 DAP = &#x2220 PCR

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 &#x2206 ABD AB2 = AD 2 + BD 2 Or AB2 = AD 2 + (BE + ED ) 2 Or AB2 = AD 2 + BE

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 &#x2206 OAR, we can write:AR2 = OA2 – OR2 —— Eqn. (I)For

Question:In the figure AB = BC and AD is perpendicular to CD. Prove that AC2 = 2.BC. DC solution :In &#x2206 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 , a line trhough A cuts DC at oint F and BC produced at E Prove that &#x2206 BCF is equal in area to &#x2206 DFE. Construction: AC is joined. AB || CD and AD || BC .Now &#x2206 ACE and &#x2206 DCE are on the same base CE and between

   ABCD is a square and APB is an equilateral triangle. Prove in each case:  1) &#x2206APD ≅ &#x2206BPC    2) Find the angles of &#x2206DPC Solution : First PartGiven &#x2206ABP = Equilateral triangleR.T.P.1) &#x2206APD ≅ &#x2206BPC    2) Find &#x2220DPC, &#x2220PDC and &#x2220PCD, Proof: In &#x2206APD and &#x2206BPC AP = BP | Same side of