Question

4. In the given figure, median AD of AABC is produced. If BL and CM are perpendiculars drawn on AD and AD produced, prove that BL = CM. [Hint. ABLD = ACMD.) B​

Answer 1

Answer:

hello,sorry for figure

inΔBDL and ΔCDM,

∠BLD=∠CMD(both equal to 90°)

∠BDL=∠CDM(vetically opposite angles are equal)

BD=CD(AD is median)

∴ΔBDL≡ΔCDM(AAS conguency)

⇒BL=CM(cpct,corresponding parts of congruent triangles)

hence proved

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Step-by-step explanation:

Answer 2

Answer:

BL⊥AD and DM⊥CM [ Given ]

In △BLD and △CMD,

⇒ ∠BLD=∠CMD=90

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[ Given ]

⇒ ∠BLD=∠MDC [ Vertically opposite angles ]

⇒ BD=DC [ Since, AD is median to BC ]

∴ △BLD≅△CMD [ By AAS congruence rule ]

We know that, corresponding parts of the congruent triangles are equal.

∴ BL=CM