Question

Given SD HT; SH=ST prove SHD= STD

Answer 1

Answer:

Δ SHD ≅ Δ STD (By SAS)

Step-by-step explanation:

Given:

SD⊥HT

SH=ST

Prove:

Δ SHD ≅ Δ STD

Computation:

SH=ST (Given)

SD = SD (Common)

∠SHD = ∠SDT (right angle) (SD⊥HT)

So,

Δ SHD ≅ Δ STD (By SAS)

Answer 2

The proof of $\Delta SHD\cong \Delta STD$ is given below.

Given, SD is perpendicular to HT.

We have to prove $\Delta SHD\cong \Delta STD$.

Now, In $\Delta SHD\ and\ \Delta STD$,

$\angle SDH=\angle SDT$ ( since SD perpendicular to HT, so both will be 90 ).

$SH=ST$ ( given in question ).

$SD= SD$ ( common side in both the triangle ).

So, by Right angle hypotenuse congruence rule,

$\Delta SHD\cong \Delta STD$.

Hence the $\Delta SHD\cong \Delta STD$ proved above.

For more details follow the link:

https://brainly.com/question/19644246