In Fig. 7.21, AC = AE, AB = AD and ∠ BAD = ∠ EAC. Show that BC = DE.

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)
Section | Number of girls per thousand boys |
|---|---|
Scheduled Caste (SC) | 940 |
Scheduled Tribe (ST) | 970 |
Non-SC/ST | 920 |
Backward districts | 950 |
Non-backward districts | 920 |
Rural | 930 |
Urban | 910 |
(i) Represent the information above by a bar graph.
(ii) In the classroom discuss what conclusions can be arrived at from the graph.
