Question

In ABC, B = 90° and DE = DC. Find EC = ?​

Answer 1

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In triangle ADB and EDC

In triangle ADB and EDCAD=DC ( Dis mid-point of AC)

In triangle ADB and EDCAD=DC ( Dis mid-point of AC)BDA = CDE ( V.O.A)

In triangle ADB and EDCAD=DC ( Dis mid-point of AC)BDA = CDE ( V.O.A)BE=DE(given)

In triangle ADB and EDCAD=DC ( Dis mid-point of AC)BDA = CDE ( V.O.A)BE=DE(given)So, ADB congurent to ECD by SAS... 1

In triangle ADB and EDCAD=DC ( Dis mid-point of AC)BDA = CDE ( V.O.A)BE=DE(given)So, ADB congurent to ECD by SAS... 1On adding triangle BDC to Triangle ADB and ECD

In triangle ADB and EDCAD=DC ( Dis mid-point of AC)BDA = CDE ( V.O.A)BE=DE(given)So, ADB congurent to ECD by SAS... 1On adding triangle BDC to Triangle ADB and ECDABC= EBC ( adding equal to equals are also equal)In triangle ADB and EDC

In triangle ADB and EDCAD=DC ( Dis mid-point of AC)BDA = CDE ( V.O.A)BE=DE(given)So, ADB congurent to ECD by SAS... 1On adding triangle BDC to Triangle ADB and ECDABC= EBC ( adding equal to equals are also equal)In triangle ADB and EDCAD=DC ( Dis mid-point of AC)

In triangle ADB and EDCAD=DC ( Dis mid-point of AC)BDA = CDE ( V.O.A)BE=DE(given)So, ADB congurent to ECD by SAS... 1On adding triangle BDC to Triangle ADB and ECDABC= EBC ( adding equal to equals are also equal)In triangle ADB and EDCAD=DC ( Dis mid-point of AC)BDA = CDE ( V.O.A)

In triangle ADB and EDCAD=DC ( Dis mid-point of AC)BDA = CDE ( V.O.A)BE=DE(given)So, ADB congurent to ECD by SAS... 1On adding triangle BDC to Triangle ADB and ECDABC= EBC ( adding equal to equals are also equal)In triangle ADB and EDCAD=DC ( Dis mid-point of AC)BDA = CDE ( V.O.A)BE=DE(given)

In triangle ADB and EDCAD=DC ( Dis mid-point of AC)BDA = CDE ( V.O.A)BE=DE(given)So, ADB congurent to ECD by SAS... 1On adding triangle BDC to Triangle ADB and ECDABC= EBC ( adding equal to equals are also equal)In triangle ADB and EDCAD=DC ( Dis mid-point of AC)BDA = CDE ( V.O.A)BE=DE(given)So, ADB congurent to ECD by SAS... 1

In triangle ADB and EDCAD=DC ( Dis mid-point of AC)BDA = CDE ( V.O.A)BE=DE(given)So, ADB congurent to ECD by SAS... 1On adding triangle BDC to Triangle ADB and ECDABC= EBC ( adding equal to equals are also equal)In triangle ADB and EDCAD=DC ( Dis mid-point of AC)BDA = CDE ( V.O.A)BE=DE(given)So, ADB congurent to ECD by SAS... 1On adding triangle BDC to Triangle ADB and ECD

In triangle ADB and EDCAD=DC ( Dis mid-point of AC)BDA = CDE ( V.O.A)BE=DE(given)So, ADB congurent to ECD by SAS... 1On adding triangle BDC to Triangle ADB and ECDABC= EBC ( adding equal to equals are also equal)In triangle ADB and EDCAD=DC ( Dis mid-point of AC)BDA = CDE ( V.O.A)BE=DE(given)So, ADB congurent to ECD by SAS... 1On adding triangle BDC to Triangle ADB and ECDABC= EBC ( adding equal to equals are also equal)

$\large\mathbb \pink{\fcolorbox{pink}{black}{ Hence, proved\ }}$

Answer 2

Answer:

Step-by-step explanation: