Question

Triangle ABD is a right triangle in which angle A =90° and AC perpendicular BD. prove that :
a)AB² = BC . BD
b)AC² =BC . DC

Answer 1

hlo mate...

❣️❤️❣️❤️❣️❤️❣️❤️❣️❤️❣️❤️


(i) In ΔADB and ΔCAB, we have
∠DAB = ∠ACB (Each equals to 90°)
∠ABD = ∠CBA (Common angle)
∴ ΔADB ~ ΔCAB [AA similarity criterion]
⇒ AB/CB = BD/AB
⇒ AB2 = CB × BD

(ii) Let ∠CAB = x
In ΔCBA,
∠CBA = 180° - 90° - x
∠CBA = 90° - x
Similarly, in ΔCAD
∠CAD = 90° - ∠CBA
= 90° - x
∠CDA = 180° - 90° - (90° - x)
∠CDA = x
In ΔCBA and ΔCAD, we have
∠CBA = ∠CAD
∠CAB = ∠CDA
∠ACB = ∠DCA (Each equals to 90°)
∴ ΔCBA ~ ΔCAD [By AAA similarity criterion]
⇒ AC/DC = BC/AC
⇒ AC2 = DC × BC

(iii) In ΔDCA and ΔDAB, we have
∠DCA = ∠DAB (Each equals to 90°)
∠CDA = ∠ADB (common angle)
∴ ΔDCA ~ ΔDAB [By AA similarity criterion]
⇒ DC/DA = DA/DA
⇒ AD2 = BD × CD


hope it HELPS ya....✌️


brainliest pls....becoz i NEED ya ..❣️✌️


@amansaini76✅

Answer 2

Answer:

DC×BC

is the correct answer