Question

with solutions please please​

Answer 1

Answer :

To Find,

• Prove ∆AEB ≅ ∆ADC

Solution,

Figure,

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,0){\line(1,0){5}}\qbezier(0,0)(0,0)(2.5,5)\qbezier(2.5,5)(2.5,5)(5,0)\qbezier(1.2,2.5)(1.2,2.5)(5,0)\qbezier(0,0)(0,0)(3.8,2.5)\put(2.3,5.3){\bf A}\put(-0.5,-0.5){\bf B}\put(5.3,-0.5){\bf C}\put(0.5,2.5){\bf D}\put(4.2,2.5){\bf E}\end{picture}$

AD = AE ...... [ Given ]

BD = CE ...... [ Given ]

AD + BD = AE + CE ..... by adding

Therefore,

BA = CA .... [ Proved ]

Proof :

BA = CA ...... [ Proved ]

AD = AE ...... [ Given ]

∠BAE = ∠CAD ...... [ Common angles ]

Therefore, ∆AEB = ∆ADC .... [ Side-angle-side property ]

Hence, Proved!

Answer 2

Step-by-step explanation:

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