AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB (see Fig). Show that
(i) ∆ DAP ≅ ∆ EBP
(ii) AD = BE
It is given that EPA = DPB
∠EPA +∠ DPE = ∠DPB + ∠DPE
∠DPA = ∠EPB
In ∆DAP and ∆EBP,
∠DAP = ∠EBP (Given)
AP = BP (P is mid-point of AB)
∠DPA = ∠EPB (From above)
∠∆DAP ∠∆EBP (ASA congruence rule)
∴ AD = BE (By CPCT)
∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that ∠ BCD is a right angle.
A driver of a car travelling at \(52\) \(km \;h^{–1}\) applies the brakes Shade the area on the graph that represents the distance travelled by the car during the period.
Which part of the graph represents uniform motion of the car?