Question 1. ABCD is a square. P is any point inside it, such that ΔPQR is another square. Prove that AP = CR
Hint: Join AP and CR.
In ΔADP and ΔCDR, we have :
AD = CD [sides of a square]
∠ADP = ∠CDR [each = 90° - ∠PDC]
DP = DR [side of a square]
⇒ ΔADP ≌ ΔCDR [SAS congruence]
⇒ AP = CR [C.P.C.T]
Question 2. E and F are the mid points of sides AB, AC of ΔABC. CE and BF are produced to X and Y respectively, such that EX = CE and FY = BF. AX and AY are joined. Find in your figure, a triangle congruent to ΔAEX and demonstrate the congruency. Show that XAY is a st. line.
Hint: Prove ΔAEX ≌ ΔBEC [By SAS congruency]
⇒ ∠ XAE = ∠ CBE [c.p.c.t.]
⇒ ∠XAB = ∠CBA
But they form a pair of co-interior angles.
⇒ XA || BC ...(1)
Similarly, ΔAFY ≌ ΔCFB
⇒ AY || BC ...(2)
from (1) and (2) XAY is a st. line.
Question 3. In the adjacent figure, BA || DF and CA || EG. If BD = EC then prove that BG = DF and EG = CF.
Hint: In ΔGBE and ΔFDC ∠ABC = ∠FDE and ∠DED = ∠ACB
also BE = DC
∴ ΔGBE ≌ ΔFDC [ASA congruency]
⇒ BG = DF and EG = CF
Question 4. ABCD is a square. M is the mid point of AB and PQ ⊥ CM meets AD at P. CB produced meet at Q. Prove that (i) PA = BQ and (ii) CP = AB + PA
Hint: Prove, ΔAMP ≌ ΔBMQ [ASA cong.]
⇒ MP = MQ and PA = QB [c.p.c.t.]
⇒ PA = BQ
Again, prove, ∆CMP ≌ ΔCMQ [SAS cong.]
⇒ CP = CQ [c.p.c.t.]
⇒ CP = CB + BQ = AB + PA